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A gaseous mixture of 3 L of propane (C_(3)H_(8)) and butane (C_(4)H_(10)) on complete combustion at 25^(@)C produced 10 L of CO_(2). Find out the composition of the gaseous mixture.

Answer» <html><body><p><br/></p>Solution :Let the volume of propane in the mixture = xL <br/> `:.` The volume of butane in the mixture `= (3-x)<a href="https://interviewquestions.tuteehub.com/tag/l-535906" style="font-weight:bold;" target="_blank" title="Click to know more about L">L</a>` <br/> Now let us calculate the volume of `CO_(2)` evolved with the help of chemical <a href="https://interviewquestions.tuteehub.com/tag/equations-452536" style="font-weight:bold;" target="_blank" title="Click to know more about EQUATIONS">EQUATIONS</a> <br/> Step I. Calculation of volume of `CO_(2)` from xL of propane <br/> The combustion equation for propane is : <br/> `underset(1L)(C_(3)H_(8))+5O_(2)rarrunderset(3L)(3CO_(2))+4H_(2)O` <br/> 1 L of propane `(C_(3)H_(8))` from `CO_(2)=3L` <br/> x L of propane `(C_(3)H_(8))` from `CO_(2)=3xL` <br/> Step <a href="https://interviewquestions.tuteehub.com/tag/ii-1036832" style="font-weight:bold;" target="_blank" title="Click to know more about II">II</a>. Calculation of volume of `CO_(2)` from (3-x)L of butane <br/> The combustion equation for butane is : <br/> `underset(1L)(C_(4)H_(10))+(13)/(2)O_(2)rarrunderset(4L)(4CO_(2))+5H_(2)O`<br/> 1L of butance `(C_(4)H_(10))` from `CO_(2)=4L` <br/> (3-x)L of butane `(C_(4)H_(10))` from `CO_(2) = 4 xx (3-x)L` <br/> Step III. Calculation of composition of the mixture <br/> Total volume of `CO_(2)` formed in the step I and step II = [<a href="https://interviewquestions.tuteehub.com/tag/3x-310805" style="font-weight:bold;" target="_blank" title="Click to know more about 3X">3X</a> + 4(3-x)]L <br/>But the volume of `CO_(2)` actually formed = 10 L <br/> `:. 3x +4(3-x)=10 , 3x +12 - 4x =10 - x = -2 or x = 2L` <br/> `:.` Volume of propane = 2L <br/> Volume of butane `= (3-2)=1L`.</body></html>


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